数组中的重复数字

public class Solution {
    // Parameters:
    //    numbers:     an array of integers
    //    length:      the length of array numbers
    //    duplication: (Output) the duplicated number in the array number,length of duplication array is 1,so using duplication[0] = ? in implementation;
    //                  Here duplication like pointor in C/C++, duplication[0] equal *duplication in C/C++
    //    这里要特别注意~返回任意重复的一个，赋值duplication[0]
    // Return value:       true if the input is valid, and there are some duplications in the array number
    //                     otherwise false
    public boolean duplicate(int numbers[],int length,int [] duplication) {
        
       
            int count=0;
            boolean f = false;
            duplication[0]=-1;
            for (int i=0;i<length;i++) {              
                for (int j=0;j<length;j++) {
                        if (numbers[i]==numbers[j]) {
                           count++;
                            if(count==2){
                            duplication[0]=numbers[j];
                            f=true;
                             break;
                            }   
                        }      
                }
                if(count==2){
                    break;
                }
                count = 0;
            }
            return f;
    }
}